So I need to test if the series $\sum_{k=1}^{\infty} (\frac{1}{k} – \frac{1}{k!})$ is absolutely convergent or not. So far, I've decided to go ahead with the ratio test, actually with a corollary of it that states that the series $\sum_{k=1}^{\infty} a_k$ is absolutely convergent if
$$\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n} \right| < 1$$
I tried to calculate it out but I'm stuck at:
$$\lim_{k\to\infty} \left( \left|\frac{k}{k+1} + \frac{k!}{k+1} – \frac{k}{(k+1)!} + \frac{k!}{(k+1)!} \right| \right)$$
I am uncertain if I can just expand the absolute value into the individual terms or leave it out completely so I can in turn calculate the limit of the individual terms too?
Any hints on how to simplify this and show the corollary above would be greatly appreciated!
Best Answer
I'd suggest using the limit comparison test with $\frac{1}{k}$ to show divergence.
If $a_k$ and $b_k$ are positive-termed sequences and $\lim_{k\rightarrow \infty} \frac{a_k}{b_k} = c$ where $c$ is positive and finite, then the series $\sum_{k=1}^\infty a_k$ and $\sum_{k=1}^\infty b_k$ either both converge or both diverge.
In this case choose $a_k = \frac{1}{k} -\frac{1}{k!}$ and $b_k = \frac{1}{k}$. Then we have \begin{align*} \lim_{k\rightarrow \infty} \frac{a_k}{b_k} & = \lim_{k\rightarrow \infty} \frac{\frac{1}{k} -\frac{1}{k!}}{\frac{1}{k}} \\ & = \lim_{k\rightarrow \infty} \left( 1-\frac{1}{(k-1)!} \right) \end{align*} It should be pretty clear that this limit is finite and positive, therefore both series diverge.